WAEC SSCE - Physics - 2023 (Essay)

Détails de la réponse

a. Fiber optics is a technology that uses light to transmit data over long distances. Fiber optic cables are made of thin, flexible strands of glass or plastic that can carry a large amount of data at very high speeds.

b. - Higher bandwidth: Optical fibers can carry a much larger amount of data than copper cables. This is because light travels much faster than electricity and optical fibers can carry more light signals than copper cables can carry electrical signals.
- Longer distances: Optical fibers can transmit data over longer distances than copper cables without signal loss. This is because light signals are less attenuated in optical fibers than electrical signals are in copper cables.
- Immunity to interference: Optical fibers are immune to electrical interference, which can cause problems with copper cables. This is because optical fibers transmit light signals, which are not affected by electrical fields.
- Security: Optical fibers are more secure than copper cables because they are difficult to tap. This is because light signals cannot be detected without damaging the optical fiber.
- Durability: Optical fibers are more durable than copper cables. They are less susceptible to damage from moisture, temperature fluctuations, and physical impact.
- Cost-effectiveness: Optical fibers are becoming more cost-effective than copper cables, especially for high-bandwidth applications. This is because the cost of manufacturing optical fibers has decreased significantly in recent years.

Détails de la réponse

ai. Electric potential at a point is defined as the work done in bringing a unit positive charge from infinity to that point against the electrical forces of the field.

ii. Process of Charging:
Induction: Involves redistributing charges within an uncharged body without direct contact with a charged object.
Contact: Requires direct physical contact with a charged object to transfer charges.
- Nature of Final Charge:
Induction: Results in a charge of opposite polarity to the charged object initially brought close.
Contact: Results in a charge of the same polarity as the charged object with which contact is made.

b. - Safety: Opening the circuit prevents current from flowing, which reduces the risk of electric shock. This is especially important when working with high voltages or currents.
- Conservation of battery power: If the circuit is left closed, the current will continue to flow even when no readings are being taken. This can drain the battery and shorten its lifespan.
- Accurate readings: If the circuit is left closed, the current flowing through the circuit can cause the components to heat up. This can affect the accuracy of the readings.
- Protection of equipment: If the circuit is left closed, a surge of current could flow through the circuit if there is a fault. This could damage the equipment or even cause a fire.

c. \(Z^2 = R^2 + X{_L}^2\)

\(Z^2 = R^2 + (2πf_L)^2

\(Z^2 = R^2 + 4π^2f^2L^2\)

\(Z^2 = 4π^2f^2L^2 + R^2\)

∴ \(Z = \sqrt{4π^2L^2f^2 + R^2}\)

ii.  from the equation

\(Z^2 = 4π^2f^2L^2 + R^2\)

using y = mx + c = y = \(Z^2\), x = \(f^2\)

m =  \(4π^2L^2\) = slope of the grape

m = \(\frac{ Z_2^2 - Z_1^2}{f_2^2 - f_1^2} = \frac{ 700 - 200 }{ 800 - 140 }\)

  \(4π^2L^2  = \frac{ 500}{ 660}\)

\(L^2 = \frac{ 500}{ 660} \times \frac{1}{ 4π^2} =\frac{ 500}{ 660} \times \frac{1}{4 \times 10}\) ( since \(π^2\) = 10)

 \(L^2 = 0.0189\) 

∴ L = \(\sqrt{ 0.0189}\) = 0.138H = 138mH

Détails de la réponse

a

i. A pivot, often referred to as a fulcrum, is the point or axis around which a simple machine, such as a lever or a seesaw, rotates.

ii. A load is the object or weight that is being moved, lifted, or supported by the machine.

iii. Efficiency refers to how effectively a machine can perform its intended task while minimizing energy loss

b.

i. Given: F = 25.2N, m = 1200kg, u = 0m/s , v = 60km/h = 16.67m/s

               Workdone = ΔK.E = \(K.E_2 - K. E_1\)

               Workdone = \(\frac{1}{2}K.E_2 - \frac{1}{2}K.E_1\)

               Workdone = \(\frac{1}{2}m(v^2 - u^2)\)

               Workdonk  =  \(\frac{1}{2}\times 1200 \times 16.67^2\)   ( since u = 0)

                 Workdone  =   600 x  277.89 = 166733.34J

                    Therefore, Workdone =  \(1.67 \times 10^5\)J

ii.  F = ma → a = \(\frac{F}{m} = \frac{25.2}{1.2 \times10^3} = 0.021ms^2\)

recall, \(v^2 = u^2 +2aS\)

 S = \(\frac{v^2 - u^2}{ 2a} = \frac{16.67^2 - 0^2}{ 2\times 0.021}\)

S = 6616.4m = 6.62km.

c. Absolute zero temperature is the lowest possible temperature that theoretically represents the complete absence of thermal energy while the ice point is the temperature at which water coexists with ice in thermal equilibrium.
- Absolute zero temperature is defined as 0 Kelvin (0 K) or approximately -273.15°C while ice point is typically defined as 0°C at sea level

d. \(\frac{0 - ( - 6)}{ 2 - L} = \frac{ 100 - 0 }{ 15 - 2}\)

\(\frac{ 6}{ 2 - L} = \frac{100 }{ 13}\)

6 x 13 = 100( 2 - L )

78 = 200 - 100L

100L = 200 - 78

100L = 122

L = \(\frac{122}{100}\)  = 1.22cm

Détails de la réponse

F = \(kv^xρ^yA^z\)
For the left-hand side:
F = mass × acceleration = \(MLT^{-2}\)
For the right-hand side:
v = \(LT^{-1}\), ρ = \(ML^{-3}\) and A = \(L^2\)
So,
\(MLT^{-2} = k(LT^{-1})^x(ML^{-3})^y(L^2)^z\)
Since k is dimensionless,
\(MLT^{-2} = (LT^{-1})^x(ML^{-3})^y(L^2)^z\)
\(MLT^{-2} = L^x × T^{(-1)x} \times M^y × L^{(-3)y} \times L^{(2)z}\)
\(MLT^{-2} = L^x × T^{-x} \times M^y \times L^{-3y} \times L^{2z}\)
\(MLT^{-2} = L^{(x - 3y + 2z)} \times T^{-x} \times M^y\)
\(M^1L^1T^{-2}= L^{(x - 3y + 2z)} \times T^{-x} \times M^{y}\)
Comparing the powers:
For M,
y = 1
For L,
x - 3y + 2z = 1 ---- (i)
For T,
-x = -2
∴ x = 2
Substitute (2) for x and (1) for y in equation (i)
⇒ 2 - 3(1) + 2z = 1
⇒ 2 - 3 + 2z = 1
⇒ -1 + 2z = 1
⇒ 2z = 1 + 1
⇒ 2z = 2
∴ z = 1

Hence, x = 2, y = 1 and z = 1.
 

Détails de la réponse

ai. The heater, often referred to as the filament, is responsible for emitting electrons when heated.

ii. The high tension source, or high voltage generator, provides the high voltage necessary to accelerate the electrons emitted from the filament towards the anode (target) of the X-ray tube.

iii. Cooling fins are designed to dissipate the heat generated during the operation of the X-ray tube.

bi. The X-ray tube is highly evacuated so that the accelerated electrons can get to their target without losing much of their energy.

ii. The target is a metal with a very high melting point to withstand the high temperature generated when the electrons strike the target.

iii. The cooling fins are placed outside the glass envelope to ensure that the heat generated within the X-ray tube is efficiently transferred to the external environment.

ci. \(^{235}U_{92} + ^1_0n →  ^{93}Kr_{36} + ^{141}Ba_{56} + 2^1_0n + Q\)

ii. Energy, Q, is released because the total mass on the left-hand side (reactant) is more than the total mass on the right-hand side (product). So, the energy is released in the form of a reduction in total mass. This missing mass is known as the 'mass defect' and it accounts for the energy released. Also, it's induced fission and nuclear fission is a process in which an unstable nucleus splits into two other lighter nuclei together with several neutrons and is accompanied by the release of energy.

iii. E = ∆\(mc^2\) where ∆m is the mass defect.
Total mass on the LHS:
235.044 u + 1.009 u = 236.053 u
Total mass on the RHS:
91.898 u + 140.914 u + 2(1.009 u) = 234.83 u
Mass defect, ∆m = 236.053 u - 234.83 u = 1.223 u
∆m = 1.223 × 1.66 × \(10^{-27}\) = 2.03 × \(10^{-27}\) kg
So,
E = 2.03 × \(10^{-27}\) × (3 × \(10^8)^2\)
= 1.827 × \(10^{-10}\) J

Détails de la réponse

ai. Simple harmonic motion is periodic because of the restoring force which is proportional to the displacement, acts in the opposite direction to the motion

ii. ♦ Length of the pendulum  ♦ Strength of the gravitational field

iii.  See the figure above.

b. m = 50g = 0.05kg, g = 10 \(ms^2\), h = 10cm = 0.1m, P.E = ?
P.E = mgh = 0.05 × 10 × 0.1
∴ P.E = 0.05 J

c. Let the extension be ∆l
F = k ∆l = k(\(l_n\) - l )
where k is the spring constant, ln is the new length, l is the natural length, and F = mg.
Total mass = mass of pan + added mass
In the first case,
\(m_1\) = 0.04 + 0.20 = 0.24 kg
\(F_1\) = 0.24 × 10 = 2.4 N
∆l = 0.055 - l
Since F = k ∆l,
2.4 = k(0.055 - l )
2.4 = 0.055k - l k --- (i)
In the second case,
\(m_2\) = 0.04 + 0.28 = 0.32 kg
\(F_2\) = 0.32 × 10 = 3.2 N
∆l = 0.065 - l
Since F = k ∆l,
3.2 = k(0.065 - l )
3.2 = 0.065k - l k --- (ii)
Subtracting equation (i) from (ii)
⇒ 0.8 = 0.01k

k = \(\frac{0.8}{ 0.01}\)

Substitute (80) for k in equation (i)
⇒ 2.4 = 0.055(80) - 80 l
⇒ 2.4 = 4.4 - 80 l
⇒ 80 l = 4.4 - 2.4
⇒ 80 l = 2

l = \(\frac{2}{80}\)
Therefore, the value of k = 80 N/m and l = 0.025 m

Détails de la réponse

  i. From v = u + gt:
\(u_x\) = ucosθ, g = 0 (horizontal component is not affected by gravity)
\(v_x\) = ucosθ + (0)t

∴   \(v_x\) = ucosθ

ii. \(v_y = u_y + gt\)
∴  \(v_y = usinθ + gt\)

b. It assumes that the only force acting on the projectile is gravity, and there are no other forces, such as air resistance or drag, affecting its motion.

Détails de la réponse

a. Strain energy is the potential energy stored within an elastic material as a result of its deformation caused by external forces.

b. Energy stored E = \(\frac{1}{2}Fe\)

Young Modulus = \(\frac{ Stress}{ Strain}\) = Y

Stress =  \(\frac{Force}{Area} = \frac{(F)}{(A)}\)

Strain =  \(\frac{ Extension}{Original Length} = \frac{(e)}{(l)}\)

Y =  \(\frac{F}{A}\div \frac{e}{l}\)

Y =  \(\frac{F}{A} \times \frac{l}{e}\)

F =  \(\frac{YAe}{l}\)

Therefore, Energy stored E = \(\frac{1}{2}\frac{YAe}{l}\times e\)

  E  =  \(\frac{YAe^2}{2l}\)